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If p+q = j and pq = k, then 2pq =
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26 Nov 2010, 13:11
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76% (01:30) correct 24% (02:15) wrong based on 329 sessions
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If p+q = j and pq = k, then 2pq = A. (j^2  k^2)/2 B. (k^2  j^2)/2 C. jk/2 D. (j^2 + k^2)/2 E. (j  k)/2 Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically... Can someone post the algebraic solution? Thanks in advance.
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Re: Solving for 2pq
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26 Nov 2010, 13:21



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Re: Solving for 2pq
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26 Nov 2010, 13:43
Thanks very much Bunuel. +1
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Re: Solving for 2pq
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26 Nov 2010, 14:03
Bunuel wrote: martie11 wrote: Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...
If p+q = j and pq = k, then 2pq =
a) (j^2  k^2)/2 b) (k^2  j^2)/2 c) jk/2 d) (j^2 + k^2)/2 e) (j  k)/2
Can someone post the algebraic solution? Thanks in advance. Given: \(p+q = j\) and \(pq = k\) > square both expressions: \(p^2+2pq+q^2=j^2\) and \(p^22pq+q^2=k^2\) > now, subtract (2) form (1): \(4pq=j^2k^2\) > \(2pq=\frac{j^2k^2}{2}\). Answer: A. Bunuel did you know to square all of the terms once you looked at the answers  since they were all in J^2 or K^2 form? I went the route of understanding this was a play on the a^2 +2ab+b^2 form but wouldn't think to subtract the two unless I looked at the answer choices. Just wanted to understand your approach.



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Re: Solving for 2pq
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26 Nov 2010, 14:23
Good question...my thought exactly...I didn't 'see' that I should square both sides...
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Re: Solving for 2pq
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27 Nov 2010, 10:49
gettinit wrote: Bunuel wrote: martie11 wrote: Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...
If p+q = j and pq = k, then 2pq =
a) (j^2  k^2)/2 b) (k^2  j^2)/2 c) jk/2 d) (j^2 + k^2)/2 e) (j  k)/2
Can someone post the algebraic solution? Thanks in advance. Given: \(p+q = j\) and \(pq = k\) > square both expressions: \(p^2+2pq+q^2=j^2\) and \(p^22pq+q^2=k^2\) > now, subtract (2) form (1): \(4pq=j^2k^2\) > \(2pq=\frac{j^2k^2}{2}\). Answer: A. Bunuel did you know to square all of the terms once you looked at the answers  since they were all in J^2 or K^2 form? I went the route of understanding this was a play on the a^2 +2ab+b^2 form but wouldn't think to subtract the two unless I looked at the answer choices. Just wanted to understand your approach. These kind of questions wud b easier if we make a note that.... (a+b)^2 = a^2 + b^2 + 2ab (ab)^2 = a^2 + b^2 2ab => a^2 + b^2 = [(a+b)^2 + (ab)^2]/2...Sum of the squares *0.5 also => 4ab = [(a+b)^2  (ab)^2]/2 ...difference of the squares *0.5



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Re: Solving for 2pq
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Updated on: 29 Nov 2010, 16:22
Look right at 2pq which hints at use of the formula (x+y)2 and (xy)2
subtracting (xy)2 from (x+y)2 above gives 4pq which when divided by 2 gives the result.
Answer: A
Originally posted by Sarang on 27 Nov 2010, 18:46.
Last edited by Sarang on 29 Nov 2010, 16:22, edited 1 time in total.



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Re: Solving for 2pq
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29 Nov 2010, 07:27
martie11 wrote: Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...
If p+q = j and pq = k, then 2pq =
a) (j^2  k^2)/2 b) (k^2  j^2)/2 c) jk/2 d) (j^2 + k^2)/2 e) (j  k)/2
Can someone post the algebraic solution? Thanks in advance. MAKE IT SIMPLE. ADD 1st equation to second ==> 2p=j+k SUBTRACT 2nd equation from 1st ==> 2q=jk multiply the above 2 equations 4pq=j^2K^2 ==> 2pq = j^2K^2 / 2 ANSWER "A" Regards, Murali. Kudos?



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Re: If p+q = j and pq = k, then 2pq =
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17 Jul 2014, 02:53
p+q = j ........... (1) pq = k ............ (2) (1) + (2) \(p = \frac{j+k}{2}\) (1)  (2) \(q = \frac{jk}{2}\) \(2pq = 2 * \frac{j+k}{2} *\frac{jk}{2}\) \(= \frac{j^2  k^2}{2}\) Answer = A
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Re: If p+q = j and pq = k, then 2pq =
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27 Oct 2016, 00:06
martie11 wrote: If p+q = j and pq = k, then 2pq =
A. (j^2  k^2)/2 B. (k^2  j^2)/2 C. jk/2 D. (j^2 + k^2)/2 E. (j  k)/2
Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...
Can someone post the algebraic solution? Thanks in advance. here you go, (p+q)^2 = p^2 + q^2 +2pq = j^2 ................1 (pq)^2 = p^2 + q^2 2pq = k^2 ..................2 subtract 2 from 1 4pq=j^2  k^2 ........................3 divide eqaution 3 by 2 on both sides 2pq = (j^2  k^2)/2 which is option A.



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Re: If p+q = j and pq = k, then 2pq =
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29 Jun 2018, 22:45
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